Question

How many grams of NaBr must be added to 250 g of water to lower the vapor pressure by 1.30 mmHg at 40 ∘C assuming complete dissociation? The vapor pressure of water at 40 ∘C is 55.3 mmHg. Can you show step by step how to do this and where everything comes from. I know this is answered on here but it's not very clear on how it's done. Thank you.

Answer #1

Moles of water = 250/18 = 13.9 moles

Lowering of vapor pressure = 1.30 mm Hg

Relative lowering of vapor pressure = 1.30/55.3 = 0.0235

Relative lowering of vapor pressure = mole fraction of NaBr

Mole fraction of NaBr = moles of NaBr/ (moles of NaBr + moles of water)

Let the moles of NaBr = x

So, relative lowering of vapor pressure = x/(x+13.9)

0.0235 = x/(x+13.9)

0.0235(x+13.9) = x

0.0235x + 0.327 = x

x = 0.327/0.9765 = 0.335

Molar mass of NaBr = 102.89 g/mol

Moles of NaBr = mass of NaBr/molar mass

0.335 = mass of NaBr/102.89

Mass of NaBr to be added = 0.335 x 102.89 = 34.47 g.

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